Let $M \subseteq \mathbb{R}^n$ be bounded and $N_{\epsilon}(M)$ the minimum number of 'squares' of side $\epsilon$ with center in M necessary to cover $M$. The box dimension of M is then defined as $\dim_B \; M = \lim\limits_{\epsilon \to 0+} \frac{\log \; N_{\epsilon}(M)}{\log \; \epsilon^{-1}}$. The following proposition shows that the s-dimensional fractal content $N_{\epsilon}(M) \epsilon^s$ presents a $0-\infty$-behaviour similar to the s-dimensional Hausdorff measure:

Proposition:1.$\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^s = \infty$ if $s < dim_B M$

2.$\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^s = 0$ if $s > dim_B M$

In argument 1, I show the existence of an element $s_0$ where the above behaviour happens. That is, I prove the proposition replacing $dim_B M$ by $s_0$.

The difference with the Hausdorff dimension definition is that in that setting we defined the dimension as the jumping value in the observed in the $0-\infty$-behaviour. Here, I need to show that $s_0 = dim_B M$.

Until now I only have an informal argument for the case $dim_B M \notin \{0,\infty\}$. So:

**Question**

How can I formally show that $s_0 = dim_B M$?

**Argument 1**

For $t > s \ge 0$ write $N_{\epsilon}(M) \epsilon^t = \epsilon^{t-s} N_{\epsilon}(M) \epsilon^s$. If $\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^s < \infty$ then $\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^t = 0$. On the other hand, if $\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^t < \infty$ and $\neq 0$ then $\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^s = \infty$. So we can confirm a $0-\infty$-behaviour as shown in the diagram:

**References:**

Falconer, Fractal Geometry: Mathematical Foundations and Applications, 2nd edition, page 46. Elgar, Measure, Topology and Fractal Geometry, 2nd edition, page 213.